Skip to main content
7 events
when toggle format what by license comment
Jul 30, 2021 at 3:41 vote accept Emily
Jul 26, 2021 at 18:03 answer added Chris Schommer-Pries timeline score: 18
Jul 26, 2021 at 12:28 comment added Toby Bartels @Denis Nardin : You're right! Then the only problem is that the link in the question is to the wrong category. (It's understandable, since the nlab doesn't have a page dedicated to the groupoid, and it is discussed as a groupoid briefly there, but somebody should add the well-known fact to it.)
Jul 25, 2021 at 8:13 comment added Denis Nardin @TobyBartels Are you sure? Note that the question is defining $\mathbb{F}$ as the groupoid of finite sets, which of I'm not mistaken is the universal symmetric monoidal $\infty$-category on one object since it's the symmetric monoidal envelope of the $\infty$-operad $Triv^\otimes$
Jul 25, 2021 at 6:07 comment added Toby Bartels A detail: $\mathbb F$ (with disjoint union $\uplus$ as the monoidal product and a singleton $\ast$ as the object) is the free symmetric monoidal category on one commutatively monoidal object. (If you take, say, $\mathbb F$ with cartesian product $\times$ and an empty set $\varnothing$ instead, then there is no monoidal functor from $(\mathbb F,\uplus)$ to $(\mathbb F,\times)$ that maps $\ast$ to $\varnothing$. The unique function from $\varnothing$ to $\ast$ would have to be mapped to a function from $\ast$ to $\varnothing$, and there is none.)
Jul 25, 2021 at 5:11 comment added John Baez Note that there's no way to generate noninvertible $n$-morphisms for $n > 0$ in the free symmetric monoidal $(\infty,\infty)$-category on one object, so it should be the same as the free symmetric monoidal $\infty$-groupoid on one object. Treated as a space, the latter should be the space of finite subsets of $\mathbb{R}^\infty$.
Jul 25, 2021 at 4:48 history asked Emily CC BY-SA 4.0